✅Pair sum (Contest)

Pair sum (Contest) easy Time Limit: 2 sec Memory Limit: 128000 kB

Problem Statement :

Given an array Arr, of N integers find the sum of max(A[i], A[j]) for all i, j such that i < j. Input The first line of the input contains an integer N, the size of the array. The second line of the input contains N integers, the elements of the array Arr.

Constraints: 1 <= N <= 100000 1 <= Arr[i] <= 100000000 Output Print a single integer which is the sum of min(A[i], A[j]) for all i, j such that i < j. Example Sample Input 1 4 5 3 3 1

Sample Output 1 24

Sample Input 2 2 1 10

Sample Output 2 10

Explanation 1 max(5,3) + max(5,3) + max(5,1) + max(3,3) + max(3,1) + max(3,1) = 24

link:https://my.newtonschool.co/playground/code/acrky37iflff/

```java
import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework

// don't change the name of this class
// you can add inner classes if needed
class Main {
    public static void main (String[] args) {
        // Your code here
        Scanner sc = new Scanner (System.in);
        int n = sc.nextInt();
        long[] arr = new long[n]; 
        for(int i = 0; i< arr.length ; i++){
            arr[i] = sc.nextInt();
        }
        
        // long max_value;
        // long sum = 0;
        // for(int i = 0; i< arr.length-1; i++){
        //     for(int j = i+1; j<arr.length ; j++){
        //         if(i<j){
        //           max_value = Math.max(arr[i], arr[j]);
        //           sum = sum + max_value;
        //         }
        //     }
        // }
        // System.out.print(sum);
        long sum = 0;
        Arrays.sort(arr);
        for(int i = 0; i< arr.length ; i++){
            sum = sum + (i*arr[i]);
        }
         System.out.print(sum);
    }
        
}
```